Two signal $x_{1} (t) a n d x_{2} (t)$ are given as $x_{1} (t) = 6 s i n c (100 t) c o s (200 t) a n d x_{2} (t) = 10 s i n c^{2} (100 t)$ . If the Nyquist sampling rate are $N_{1}$ and $N_{2}$ respectively, then the value of $\frac{N_{1}}{N_{2}}$ is ______(Upto 1 decimal place)

1.5

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Solution

The correct option is A 1.5
$x_{1} (t) = 6 s i n c (100 t) c o s (200 π t)$

$X_{1} (f)$ is convolution of two signals whose spectrum covers $f = \pm 50 H z$ and $\pm 100 H z$ . So convolution extends over $f = 150 H z$ and sampling rate $N_{1} = 2 f = 300 H z$

$x_{2} (t) = 10 s i n c^{2} (100 t)$

Taking Fourier transform

$X_{2} (f) = 0.1 t r i (\frac{f}{100})$

$f_{m a x} = 100 H z$

$N_{2} = 2 f_{m a x} = 200 H z$

$∴ \frac{N_{1}}{N_{2}} = \frac{300}{200} = 1.5$

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